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If I have a continued fraction for an irrational number $z= \langle a_0;a_1,a_2,a_3,\ldots\rangle$ it seems that $(-1)*z = \langle-a_0;-a_1,-a_2,-a_3,\ldots\rangle$. Is this true?

In general, if you have the continued fraction representation for $y$ and $z$ can you say something about the continued fraction representation of $y*z$?

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  • $\begingroup$ I fixed your LaTeX. Note that you should include entire formulas in between dollar signs, not parts of them: write $z= \langle a_0;a_1,a_2,a_3,\ldots\rangle$. Also, due to html you should not use < and >, rather use $\lt$ and $\gt$. $\endgroup$
    – t.b.
    Nov 15, 2011 at 8:26
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    $\begingroup$ Maybe you should give a definition of the continued fraction representation. $\endgroup$
    – Phira
    Nov 15, 2011 at 8:31
  • $\begingroup$ inwap.com/pdp10/hbaker/hakmem/cf.html $\endgroup$
    – Peter Taylor
    Nov 15, 2011 at 8:44
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    $\begingroup$ Phira is probably hinting to you that "continued fraction" is commonly interpreted to mean "regular continued fraction" and that means that $a_1,a_2,\dots$ must all be positive, so your fraction for $-z$ doesn't qualify. $\endgroup$
    – Gerry Myerson
    Nov 15, 2011 at 12:04
  • $\begingroup$ I worked out a formula for the "correct" continued fraction of $-z$ (using positive integer entries after the first term) at kconrad.math.uconn.edu/blurbs/ugradnumthy/…. It involves two cases, depending on whether the second entry is $1$ or is $\geq 2$, assuming $z$ is not an integer = 1-term continued fraction (in my write-up, I start indexing with $a_1$, not $a_0$, so your $a_1$ is my $a_2$). The continued fraction of $1/z$ is worked out there as well, and it involves 10 cases. $\endgroup$
    – KCd
    Nov 1, 2020 at 1:31

2 Answers 2

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While evaluating the continued fraction of the negative coefficients of the continued fraction expansion of $z$ does indeed evaluate to $-z$, your formula $[-a_0; -a_1, \dots, ]$ is not regarded as "the continued fraction" of $-z$, which is usually defined using the Euclidean algorithm resulting in non-negative coefficients after the first. In general, for $z \in \mathbb{R}$ and $z= [a_1; a_2, a_3 \dots]$, then \begin{align} -z= [-a_1-1;1,a_2-1, a_3, \dots], \end{align} where the terms in the ellipses are identical. For example, $\frac{4}{3} = [1,3]$, while $-\frac{4}{3} = [-2,1,2]$. To address your second question, there are formulas to compute the continued fraction expansion of $\frac{az+b}{cz+d}$, where $a, b, c, d \in \mathbb{Z}$, relying only on the continued fraction expansion of $z$ and certain $2 \times 2$ matrices defined using the coefficients. See An Introduction to Continued Fractions by van der Poorten.

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  • $\begingroup$ Which notes by Pomerance? $\endgroup$
    – Igor Rivin
    Dec 23, 2013 at 17:31
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    $\begingroup$ Your formula for $-z$ is not always correct. For example, $\sqrt{3} = [1;\overline{1,2}]$, and $-\sqrt{3} = [-2; 3, \overline{1,2}]$. If $z= [a_0; 1, a_2, a_3, a_4, \dots]$ is irrational, then $-z = [-a_0 - 1; a_2 + 1, a_3, a_4, \dots]$. Your formula is good if the second partial quotient of $z$ is greater than 1. $\endgroup$
    – user0
    May 16, 2017 at 16:19
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    $\begingroup$ @user0 Just for the record the given $-z$ formula is not totally wrong however what you mention is a special case. As you say if $a_1=1$ then we will end up like $[-a_0;1,0,a_2,...,a_n]$. So regardless of negation, whenever we find a $0$ within the CF coefficients we can safely merge the previous and next items by adding, in this case yielding $[-a_0;1+a_2,...,a_n]$ as you have shown in your comment. $\endgroup$
    – Redu
    May 12, 2022 at 15:55
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If continued fraction $$ a_0+\frac{1}{\displaystyle a_1+\frac{1}{a_2+\ddots}} $$ converges to $z$, where all $a_k$ and $z$ are complex numbers, then continued fraction $$ -a_0+\frac{1}{\displaystyle -a_1+\frac{1}{-a_2+\ddots}} $$ converges to $-z$.

Is that what you mean? You should be able to prove it!

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